Is the space $\mathcal C([0,1])$ endowed with the sup norm homeomorphic to $\mathcal C([0,1])$ endowed with the integral norm?

Solution 1:

If we embed $(C(X),\|\cdot\|_1)$ into its Banach space completion $E$, then if $C(X)$ were a $G_\delta$ in $E$ then any $f \in E\setminus C(X)$ would contradict Baire's theorem for $E$ as $f + C(X)$ and $C(X)$ are disjoint dense $G_\delta$'s in $E$. So in fact $E=C(X)$. (An old argument by Mazur).

Conclusion: $(C(X),\|\cdot\|_1)$ is not topologically complete (aka Čech-complete). So it cannot be homeomorphic to $(C(X),\|\cdot\|_\infty)$, which is.

The Kadec-Anderson theorem says that if $X$ is a completely metrisable, separable locally convex TVS, then $X \simeq \Bbb R^\omega \simeq \ell_2$, topologically. So the non-completeness is the obstacle.